Question 288851
{{{n^(n-1)-1}}} is divisible by {{{(n-1)^2}}} whenever {{{n>=2}}}
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Lemma: there exists non-negative integer q such that

{{{n^k/(n-1)=q+1/(n-1)}}}

Proof:

By a factorization theorem {{{n^k-1}}} can be factored as:

{{{n^k-1=(n-1)(n^(k-1)+n^(k-2)+"..."+n^2+n+1)}}}

where there are k terms in the second parentheses

{{{n^k=(n-1)(n^(k-1)+n^(k-2)+"..."+n^2+n+1)+1}}}

Therefore {{{n^k}}}{{{mod}}}{{{(n-1)}}}{{{"="}}}{{{1}}}

or there exists positive integer q such that

{{{n^k/(n-1)=q+1/(n-1)}}}

Thus the lemma is proved.

{{{n^(n-1)=(n-1)(n^(n-2)+n^(n-3)+"..."+n^2+n+1)}}}

where there are n-1 terms in the parentheses on the right.

Now we wish to show that

{{{(n^(n-1)-1)/(n-1)^2}}} is a positive integer.  Factoring the numerator:

{{{(n^(n-1)-1)/(n-1)^2=((n-1)(n^(n-2)+n^(n-3)+"..."+n^2+n+1))/(n-1)^2=""}}}

{{{(cross((n-1))(n^(n-2)+n^(n-3)+"..."+n^2+n+1))/(n-1)^cross(2)=""}}}

{{{(n^(n-2)+n^(n-3)+"..."+n^2+n+1)/(n-1)=""}}}

{{{(n^(n-2))/(n-1)+(n^(n-3))/(n-1)+"..."+n^2/(n-1)+n/(n-1)+1/(n-1)=""}}}

By the lemma, there exist {{{q[1]}}},{{{q[2]}}},...{{{q[n-1]-""}}} so
that the preceding expression equals

{{{(q[1]+1/(n-1)) + (q[2]+1/(n-1)) + (q[3]+1/(n-1)) + "...+("}}}{{{q[n-1]-""}}}{{{""+1/(n-1))}}}{{{")="}}}

{{{"("}}}{{{q[1]+q[2]+q[3]+"..."+q[n-1]-")+"}}}{{{1/(n-1)+1/(n-1)+"..."+1/(n-1)}}}

since there are {{{n-1}}} terms this becomes:

{{{"("}}}{{{q[1]+q[2]+q[3]+"..."+q[n-1]-")+"}}}{{{(n-1)/(n-1)}}}

or

{{{q[1]+q[2]+q[3]+"..."+q[n-1]-""}}}{{{""+1}}},

which is a positive integer.

Edwin</pre>